Sub Topics

To describe the motion of a body in circular path is very easy to understand using angular quantities. This angular quantities will also affect the linear quantities of a body in circular motion. There are two direction of motion when we considering any point on a circle. One is directed away from the center that is along the radius which is known as radial direction and the other one is perpendicular to that which is known as tangential direction. A tangent means a line which just pass through one point on a circle and makes a right angle with the radius of the circle. When we considering a circular motion of an object, we need to consider the changes in the direction and magnitude of the velocity. The direction of velocity at any point on the circular path is same as that of the tangent. Whereas the direction of acceleration is not same as that of the tangent. The more details about the tangential acceleration that is definition, equation etc. are described in following sections. 
The tangential acceleration is defined as the rate of change of tangential velocity with time. And in fact, the relationship between tangential acceleration and angular acceleration is the same as the relationship between tangential velocity and angular velocity. In another words, the tangential acceleration is the instantaneous linear acceleration of a body directed to the tangent of the body's circular path.
$a_{t}$ = r $ \alpha $
Where a_{t} is the tangential accelerationr is the radius
α is the angular acceleration
Or it is represented as
$a_{t}$ = $ \frac{dv}{dt} $
Where dv is change in angular velocitydt is change in time
The SI unit of tangential acceleration (a_{t}) is m/s^{2}, which is same as that of normal acceleration. But the unit of radius (r) is m and angular acceleration (α) is rad/s^{2}.
Before explaining about the tangential acceleration, we should know about the direction along tangent. Consider a circle in which a particle is moving in circular direction. In this circular motion, two directions are possible for a particle. One is tangential and the other is radial direction. Radial direction is directed away from the center and the tangential direction is perpendicular to the radial direction. So, the acceleration of a particle in circular motion has both tangential as well as radial components. Radial component is also known as the normal component. The direction of the tangential component is same as that of the direction of the tangent itself. The tangential component of acceleration denoted by a_{t}.
Tangential speed is a measure of linear speed of a particle which is under a circular motion. This can be denoted as Vt and the formula is given by V_{t} = rω
Where V_{t} is the tangential speedr is the radius of the circular path
ω is the angular velocity
Example:
Solved Example
Question: Determine the tangential speed of a particle which is in a circular motion, if the radius of the curvature is given as 3 m and the time is 5 s?
Solution:
The given parameters are
r = 3 m; t = 5 s
We know that, $ \omega $ = $ \frac{2\pi }{t} $
So, $V_{t}$ = r $ \omega $
$V_{t}$ = 3 $ \times$ $ \frac{2\times 3.14}{5} $
$V_{t}$ = 3.768 m/s
Solution:
The given parameters are
r = 3 m; t = 5 s
We know that, $ \omega $ = $ \frac{2\pi }{t} $
So, $V_{t}$ = r $ \omega $
$V_{t}$ = 3 $ \times$ $ \frac{2\times 3.14}{5} $
$V_{t}$ = 3.768 m/s
Tangential and normal acceleration components are two perpendicular components of a particle which is under a circular motion. The idea about the tangential component is mentioned above. Normal component is also known as the radial component. The tangential acceleration is due to the tangential force whereas the radial acceleration is due to the centripetal force. Which is acting towards the center. The resultant acceleration will be the vector sum of radial component and the tangential component of the acceleration.
The formulas for tangential and radial acceleration is given byTangential acceleration:
$a_{t}$ = r $ \alpha $
Where a_{t} is the tangential acceleration
r is the radius
α is the angular acceleration
Radial acceleration:
$a_{r}$ = $ \frac{v^{2}}{r} $
Where a_{r} is the radial acceleration
v is the velocity
r is the radius of curvature
Radial acceleration is due to the centripetal force, hence it is also known as the centripetal acceleration.$a_{t}$ = r $ \alpha $
Where a_{t} is the tangential acceleration
r is the radius
α is the angular acceleration
Radial acceleration:
$a_{r}$ = $ \frac{v^{2}}{r} $
Where a_{r} is the radial acceleration
v is the velocity
r is the radius of curvature
Angular and tangential accelerations are inter connected. angular acceleration is defined as the rate of change of angular velocity with time. Whereas the tangential acceleration is given by the rate of change of tangential velocity with time. The equations of angular and tangential acceleration is given below:
Tangential Acceleration:
$a_{t}$ = r $ \alpha $
Angular Acceleration:
$ \alpha $ = $ \frac{a_{t}}{r} $
Where a_{t} is the tangential acceleration
r is the radius
α is the angular acceleration
$a_{t}$ = r $ \alpha $
Angular Acceleration:
$ \alpha $ = $ \frac{a_{t}}{r} $
Where a_{t} is the tangential acceleration
r is the radius
α is the angular acceleration
In some cases angular velocity and tangential velocity may differ. For a clear understanding, just consider two points on a circle. One is 1 meter far from the axis of rotation and other is 2 meters away from the axis. When it starts to rotate, both of them have same angular speed since the angle between the initial and final positions are same. But they have the different tangential speeds. If the point is away from the center it moves faster. In order to maintain the equal angular displacement, the longer radius point must travel through a larger arc with in the same time.
Problem related to this topic is given below:
Solved Examples
Question 1: Calculate the tangential acceleration of a car which accelerates uniformly on a circular direction. The velocity is changing from 30 m/s to 60 m/s with in 10 s?
Solution:
From the question it is clear that,
Initial velocity, v_{i} = 30 m/s ; final velocity, v_{f} = 60 m/s ; time, t = 10 s
We know that,
$a_{t}$ = $ \frac{dv}{dt} $
Change in velocity, dv = v_{f}  v_{i} = 60  30 = 30 m/s
Time taken, dt = 10 s
So, $a_{t}$ = $ \frac{30}{10} $
$a_{t}$ = 3 m/s^{2}
Solution:
From the question it is clear that,
Initial velocity, v_{i} = 30 m/s ; final velocity, v_{f} = 60 m/s ; time, t = 10 s
We know that,
$a_{t}$ = $ \frac{dv}{dt} $
Change in velocity, dv = v_{f}  v_{i} = 60  30 = 30 m/s
Time taken, dt = 10 s
So, $a_{t}$ = $ \frac{30}{10} $
$a_{t}$ = 3 m/s^{2}
Question 2: Consider a bug on the rim of a 0.2 m radius disc. If the disc moves from rest to an angular speed of 78 rev/min with in 3 s, determine the tangential acceleration of this bug?
Solution:
Given that, t = 3 s, r = 0.2m
The angular velocity is given by,
$ \omega $ = 78 $\frac{rev}{min}$ $\times$ $\frac{2 \pi rad}{1rev}$ $\times$ $\frac{1min}{60s}$ = 8.2 rad/s
The angular acceleration of the disc is given by,
$ \alpha $ = $ \frac{\omega \omega _{0}}{t} $
$ \alpha $ = $ \frac{8.20}{3} $ = 2.7 rad/s^{2}
So the tangential acceleration of a bug is,
$a_{t}$ = r $\alpha$
$a_{t}$ = 0.2 $\times$ 2.7 = 0.54 $m/s^{2}$
Solution:
Given that, t = 3 s, r = 0.2m
The angular velocity is given by,
$ \omega $ = 78 $\frac{rev}{min}$ $\times$ $\frac{2 \pi rad}{1rev}$ $\times$ $\frac{1min}{60s}$ = 8.2 rad/s
The angular acceleration of the disc is given by,
$ \alpha $ = $ \frac{\omega \omega _{0}}{t} $
$ \alpha $ = $ \frac{8.20}{3} $ = 2.7 rad/s^{2}
So the tangential acceleration of a bug is,
$a_{t}$ = r $\alpha$
$a_{t}$ = 0.2 $\times$ 2.7 = 0.54 $m/s^{2}$